Prove...

$\displaystyle \frac{1}{sec(x)-tan(x)}=sec(x)+tan(x)$

this is as far as i got

working from LHS

$\displaystyle \frac{1}{sec(x)-tan(x)}$

$\displaystyle cos(x)-cot(x)$

$\displaystyle cos(x)-\frac{cos(x)}{sin(x)}$

$\displaystyle \frac{sin(x)}{sin(x)}cos(x)-\frac{cos(x)}{sin(x)}$

$\displaystyle \frac{sin(x)cos(x)}{sin(x)}-\frac{cos(x)}{sin(x)}$

now i'm stuck..